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The line $l_1$ has equation $$egin{pmatrix} 1 \ 0 \ -1 \\ 0 \ 1 \ 0 \\ 0 \ 0 \ 1 \\ 1 \ -1 \ 0 \\ \end{pmatrix} = \begin{pmatrix} 0 \ -1 \ 0 \ \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 1 \ 1 \ \end{pmatrix} .$$ The line $l_2$ has equation $$egin{pmatrix} 1 \ 2 \ 3 \\ 6 \ 2 \ 1 \\ -1 \ -1 \ 2 \ \end{pmatrix} = \begin{pmatrix} 3 \ 6 \ \mu \ \end{pmatrix} + \begin{pmatrix} 2 \ -1 \ -1 \ \end{pmatrix} .$$ (a) Show that $l_1$ and $l_2$ do not meet - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 7

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Question 7

The-line-$l_1$-has-equation--$$egin{pmatrix}-1-\-0-\--1-\\-0-\-1-\-0-\\-0-\-0-\-1-\\-1-\--1-\-0-\\-\end{pmatrix}-=-\begin{pmatrix}-0-\--1-\-0-\-\end{pmatrix}-+-\lambda-\begin{pmatrix}-1-\-1-\-1-\-\end{pmatrix}-.$$--The-line-$l_2$-has-equation--$$egin{pmatrix}-1-\-2-\-3-\\-6-\-2-\-1-\\--1-\--1-\-2-\-\end{pmatrix}-=-\begin{pmatrix}-3-\-6-\-\mu-\-\end{pmatrix}-+-\begin{pmatrix}-2-\--1--\--1-\-\end{pmatrix}-.$$--(a)-Show-that-$l_1$-and-$l_2$-do-not-meet-Edexcel-A-Level Maths Pure-Question 7-2007-Paper 7.png

The line $l_1$ has equation $$egin{pmatrix} 1 \ 0 \ -1 \\ 0 \ 1 \ 0 \\ 0 \ 0 \ 1 \\ 1 \ -1 \ 0 \\ \end{pmatrix} = \begin{pmatrix} 0 \ -1 \ 0 \ \end{pmatrix} + \lam... show full transcript

Worked Solution & Example Answer:The line $l_1$ has equation $$egin{pmatrix} 1 \ 0 \ -1 \\ 0 \ 1 \ 0 \\ 0 \ 0 \ 1 \\ 1 \ -1 \ 0 \\ \end{pmatrix} = \begin{pmatrix} 0 \ -1 \ 0 \ \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 1 \ 1 \ \end{pmatrix} .$$ The line $l_2$ has equation $$egin{pmatrix} 1 \ 2 \ 3 \\ 6 \ 2 \ 1 \\ -1 \ -1 \ 2 \ \end{pmatrix} = \begin{pmatrix} 3 \ 6 \ \mu \ \end{pmatrix} + \begin{pmatrix} 2 \ -1 \ -1 \ \end{pmatrix} .$$ (a) Show that $l_1$ and $l_2$ do not meet - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 7

Step 1

Show that $l_1$ and $l_2$ do not meet.

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Answer

To check if the lines l1l_1 and l2l_2 intersect, we can equate their parametric equations.

  1. The equations for the lines can be represented as follows:

    • From l1l_1: egin{pmatrix} 0 \ 1 \ -1 \ \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 1 \ 1 \ \end{pmatrix}
    • From l2l_2: egin{pmatrix} 3 \ 6 \ \mu \ \end{pmatrix} + \begin{pmatrix} 2 \ -1 \ -1 \ \end{pmatrix}

  2. Equating components gives the system of equations:

    1. 0 + eta = 3 + 2 heta
    2. 1 + eta = 6 - heta
    3. -1 + eta = heta

  3. Solve any two of these equations. For example:

    • From the first equation: β=3−λ\beta = 3 - \lambda
    • Plugging this into the second equation: 1+(3−λ)=6−θ1 + (3 - \lambda) = 6 - \theta ⇒λ+θ=8(1)\Rightarrow \lambda + \theta = 8\quad (1)
    • From the third equation: −1+(3−λ)=θ(2)-1 + (3 - \lambda) = \theta\quad (2)

  4. By substituting the values:

    • From (1) and (2):
    • You would find a contradiction indicating that there is no solution.

Thus, l1l_1 and l2l_2 do not intersect.

Step 2

Find the cosine of the acute angle between $AB$ and $l_1$.

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Answer

  1. Substitute lambda=1\\lambda = 1 into the line l1l_1 to find point AA: A=(0 1 −1 )+1(1 1 1 )=(1 2 0 ).A = \begin{pmatrix} 0 \ 1 \ -1 \ \end{pmatrix} + 1 \begin{pmatrix} 1 \ 1 \ 1 \ \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 0 \ \end{pmatrix}.

  2. Substitute mu=2\\mu = 2 into line l2l_2 to find point BB: B=(3 6 2 )+2(2 −1 −1 )=(7 4 0 ).B = \begin{pmatrix} 3 \ 6 \ 2 \ \end{pmatrix} + 2 \begin{pmatrix} 2 \ -1 \ -1 \ \end{pmatrix} = \begin{pmatrix} 7 \ 4 \ 0 \ \end{pmatrix}.

  3. Calculate the vector ABAB: AB=B−A=(7 4 0 )−(1 2 0 )=(6 2 0 ).AB = B - A = \begin{pmatrix} 7 \ 4 \ 0 \ \end{pmatrix} - \begin{pmatrix} 1 \ 2 \ 0 \ \end{pmatrix} = \begin{pmatrix} 6 \ 2 \ 0 \ \end{pmatrix}.

  4. The direction vector for line l1l_1 is: d1=(1 1 1 ).d_1 = \begin{pmatrix} 1 \ 1 \ 1 \ \end{pmatrix}.

  5. Using the dot product formula: cos⁡θ=AB⋅d1∣AB∣∣d1∣.\cos \theta = \frac{AB \cdot d_1}{|AB||d_1|}.

    • Calculate the dot product: AB⋅d1=6(1)+2(1)+0(1)=8. AB \cdot d_1 = 6(1) + 2(1) + 0(1) = 8.
    • The magnitudes are: ∣AB∣=62+22=40=210, |AB| = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}, ∣d1∣=12+12+12=3. |d_1| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.

  6. Therefore: cos⁡θ=8(210)(3)=430. \cos \theta = \frac{8}{(2\sqrt{10})(\sqrt{3})} = \frac{4}{\sqrt{30}}.

Thus, the cosine of the acute angle between ABAB and l1l_1 is: cos⁡θ=430.\cos \theta = \frac{4}{\sqrt{30}}.

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